What is the solution for y^'=(x+1)(3y-1)2 ?

The solution for y^'=(x+1)(3y-1)2 is y=(e^{3x^2+6x+c_{1}})/3+1/3

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y^'=(x+1)(3y-1)2

y^{^{'}} = (x + 1) (3 y - 1) 2

y = \frac{e ^{3 x^{2} + 6 x + c_{1}}}{3} + \frac{1}{3}

Solution steps

y^{'} (x) = (x + 1) (3 y - 1) 2

Solve separable ODE:

y = \frac{e ^{3 x^{2} + 6 x + c_{1}}}{3} + \frac{1}{3}

y = \frac{e ^{3 x^{2} + 6 x + c_{1}}}{3} + \frac{1}{3}

\frac{d}{d x} (\frac{8 x + x ^{2}}{( 4 + x ) ^{2}})

f (x) = 13 x^{3} (x^{2} - 3)

y = \frac{x}{81 + x ^{2}}, (0, 0)

\int \frac{14}{x ( x ^{2} + 25 )} d x

y^{^{''}} - 4 y^{^{'}} - 5 y = 0

What is the solution for y^'=(x+1)(3y-1)2 ?
The solution for y^'=(x+1)(3y-1)2 is y=(e^{3x^2+6x+c_{1}})/3+1/3